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November 28th, 2001, 06:03 AM
#1
**Warning: Math**
(I have to much time on my hands)
Currently, n divided by zero(0) is undefined.
Why? What are your thoughts on this, and any possible theories for a definition for n/0 ?
********My Theories********
Code:
Example 1.1:
1 | 2 | 3
5 * 6 = 30 | 30 / 5 = 6 | 30 / 6 = 5
10 * 3 = 30 | 30 / 10 = 3 | 30 / 3 = 10
Example 1.2:
1 | 2 | 3
0 * 5 = 0 | 0 / 0 = ? | 0 / 5 = ?
0 * 10 = 0 | 0 / 0 = ? | 0 / 10 = ?
In example 1.1, If two numbers are multiplied together, their
product divided by one of the numbers should only equal
the other number. This is because the only way to reach the
product with the given number is to multiply it(or add it to itself)
x amount of times, where x is the other number.
Of course, the same should apply to division by zero, and if
you think about, it indeed does in column three of example 1.2.
Zero divided by zero, however, could be any given number,
because any number multiplied by zero equals zero.
This is where one of the problems of defining division by zero
arises. In one case, where zero is divided by a non-zero number,
It follows basic rules, but in the other, zero divided by zero, it is
impossible to tell what it should be. What could you put in there
to make the statement always correct?
Now for the next part:
Example 2.1:
2 / 0 = ?
10 / 0 = ?
The same as the zero divided by zero case, except with a new
twist: zero multiplied by anything will never equal a number
other than zero, but now we're trying to divide a number by
another number that would never multiply into it?! That the
denominator should multiply into the numerator is essential in the
multiplication/division relationships!
Since it is impossible to achieve an answer through the former
methods of thought, I propose a new one. In division, you
basically take the numerator and put equal amounts of it into a
number of groups equal to the denominator and finding out the
number in any given group.
Example 3.1:
10 put equally into 5 groups = 2 in any group
30 put equally into 3 groups = 10 in any group
Following this method, (this could be a bit confusing: )putting
a number into zero groups should equal zero in any of the group
(The confusing bit: even though there are no groups..), so:
Example 3.2:
10 put equally into 0 groups = 0 in any group
30 put equally into 0 groups = 0 in any group
Problem: what happened to the initial amount? If this were
done with energy, which can be neither created nor destroyed,
this would be impossible. So now I propose, in conjunction with
this, that a special convention be applied to this case.
A convention such as I mentioned has already been entered
into algebraic math: A negative number has no real square
root, but this has been solved by putting an i on the outside
of the square in this way:
Example 4.1:
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i
etc.
Example 4.2:
root -16 = root (16 * i^2) = i(root 16) = 4i
root -49 = root (49 * i^2) = i(root 49) = 7i
2i = 2i
2i^2 = -2
2i^3 = 2(-i)
2i^4 = 2
Use of a convention such as this would definitely help in the
continuance of a problem without a stall at the point of division
by zero. How to implement this convention presents yet another
problem, but which I, at this time, unfortunately have no idea
how to do.
If you have gotten this far, thank you for your time, if you understood and could follow me, even better. Either way, all comments, thoughts, or theories would be much appreciated. Again, thanks for your time and input.
Preliminary operational tests were inconclusive (the dang thing blew up)
\"Ask not what the kernel can do for you, ask what you can do for the kernel!\"
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November 28th, 2001, 08:53 AM
#2
Here's my opinion... Just follow the trend! If you use:
y = 1/X
As X increases, Y decreases towards a limit at zero.
As X gets closer to zero, y becomes closer to ±infinity (depending on sign (±) of X)
Therefore it makes perfect sense to define
Constant Number / 0 = Infinity
Now, computers don't do well with infinite values, hence the 'Divide by Zero' error. It's just not a safe or predictable thing for most programs to ever attempt.
Now, if you look at the other case:
y = 0/X
No matter WHAT you do, the limit of the function is zero, so it clearly points to y being zero at all times, meaning 0/0 is 0.
Summary:
0/0 is 0
A/0 is ±infinity, where A is a nonzero number
It's really simple in calculus.
[HvC]Terr: L33T Technical Proficiency
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November 28th, 2001, 09:43 AM
#3
Senior Member
It's simple to visualize if you take apples and oranges.
Take 30 apples and seperate them into three groups, the groups will be of size 10, right? (ie, 30/3=10)
Ok, now take those same 30 apples and seperate them into 0 groups. Impossible to do! (ie. 30/0 is undefined)
Logic wins again.
You encounter this stuff all the time such as when taking derivatives of functions like f(x) = 1/(x*x - 1) = 1/[(x-1)(x+1)]
Therefore when using this function for anything x may not equal + or - 1.
Anyways, I've done way too much calculus. Doh, just remembered I have a midterm on friday that includes laplace transformations (don't you just love those guys!) and power series solutions to differential equations. Needless to say i should go study.
- Stronzo
\"Vini, Vici, Vidi\"
I came, I saw, I conquered.
- Julius Caesar
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November 28th, 2001, 12:24 PM
#4
hehe
every body knows anything diveded by 0 = 0 sorry just remebering my yr 10 maths class when some one said that ohh we laft or is y=1/2x2 half a porabla hehe
sorry im babaling
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November 28th, 2001, 02:09 PM
#5
Originally posted by Stronzo
It's simple to visualize if you take apples and oranges.
Take 30 apples and seperate them into three groups, the groups will be of size 10, right? (ie, 30/3=10)
Ok, now take those same 30 apples and seperate them into 0 groups. Impossible to do! (ie. 30/0 is undefined)
Logic wins again.
Not really. Logic only wins again when you realise that you've still got 1 group of 30 apples(Assuming you're working from the start again), and that nothing has changed.
You can do the same with a pie. Cut a pie into 0 pieces (dividing by 0), which is impossible, so you are left with whatever you started.
Chris Shepherd
The Nelson-Shepherd cutoff: The point at which you realise someone is an idiot while trying to help them.
\"Well as far as the spelling, I speak fluently both your native languages. Do you even can try spell mine ?\" -- Failed Insult
Is your whole family retarded, or did they just catch it from you?
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November 28th, 2001, 03:09 PM
#6
Originally posted by Kezil
Currently, n divided by zero(0) is undefined.
Why? What are your thoughts on this, and any possible theories for a definition for n/0 ?
n divided by zero indeed is undefined, and it should stay that way
Why? Because if you allow a 'Divide by Zero', you'll get things like this:
Code:
Premise: a = b
1. a²=ab
Because a = b
2. a² + a² = a² + ab
Add a²
3. 2a² = a² + ab
x + x = 2x
4. 2a² - 2ab = a² + ab - 2ab
Add -2ab
5. 2a² - 2ab = a² - ab
xy - 2xy = -xy
6. 2(a² - ab) = 1(a² - ab)
Single out the factors
7. {2(a² - ab)} / {a² - ab} = {1(a² - ab)} / {a² - ab}
I'll get to this one in a few...
2 = 1
There you go: 2 = 1
A practical example:
Code:
Premise: a=2 and b=2
1. 2² = 2 x 2
2. 4 + 4 = 4 + 4
3. 2 x 4 = 4 + 4
4. 2 x 4 - 8 = 4 + 4 - 8
5. 2 x 4 - 2 x 4 = 4 - 4
6. 2(4 - 4) = 1(4 - 4)
Steps 1 - 6 are correct, but step 7 is why x / 0 should be undefined:
Code:
2 / (4 - 4) = 1 / (4 - 4)
thus proving that 1 = 2
Damn, I love Pythagoras...
x / 0 is not infinity, x / 0 is undefined, and you should be glad it is
Originally posted by Terr
Summary:
0/0 is 0
A/0 is ±infinity, where A is a nonzero number
Summary:
0/0 is undefined
A/0 is undefined, where A is whatever (apples, oranges, pie,...)
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November 28th, 2001, 03:53 PM
#7
Junior Member
in defense of imaginary numbers
Well, first I have to agree with Negative that anything divided by zero is undefined. As such you should simply take the limit as the value in the denominator approaches zero, and use l'Hopital's rule if necessary.
The convention suggested is an interesting idea but I'm not sure how good an idea it is to simply create conventions so late in the mathematical game. The use of i to represent the square root of negative one is really no different that using pi to represent the relationship between the diameter and circumference of a circle. The fact that numbers containing i have sometimes been called "imaginary" is just unfortunate nomenclature as they are just as real and important as the number set with which most people are comfortable.
I myself was not a fan of the complex plane until recently. However, once you realize that all numbers are complex (some just have a zero-size imaginary component) higher-level math makes a lot more sense, and though it may make your life more complicated, you'll have a much better understanding of physical and electical systems.
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November 28th, 2001, 04:50 PM
#8
Originally posted by Terr
Now, computers don't do well with infinite values, hence the 'Divide by Zero' error. It's just not a safe or predictable thing for most programs to ever attempt.
I propose a solution for the whole Divide by Zero cunnundrum (pardon spelling)...
Introduce a third binary digit.. The Trinary Bit (using the same method for the previous bit type item, its acronym is TiT)...
This would makes the computers logic have three choice..
Yes (On)
No (Off)
Maybe(The TiT)
Anyway, I'm drunk and have now lowered your intellegence 20%.. enjoy..
-Matty_Cross
\"Isn\'t sanity just a one trick pony anyway? I mean, all you get is one trick. Rational Thinking.
But when you\'re good and crazy, hehe, the skies the limit!!\"
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November 28th, 2001, 05:52 PM
#9
Member
Cool post, I am a math geek as well. I am curious if a cryptography section would be a good idea here on AO. I realize that it may only be Kezil and the people who repled to Kezil's post that would be interested. Just a thought.
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November 28th, 2001, 06:04 PM
#10
psi0nic: two words for you: Private forums ! JP promised me yesterday he'd have them up by the end of the day . Well, they're not up yet, but I guess they'll be soon...
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